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Correct formula to use for potential difference?

  1. Sep 14, 2013 #1
    My book can't stick to one formula EVER.
    In Q7 They start off using: E = −ΔV Δs (equation 20-4)

    Then magically go to: ΔV = EΔs

    Where is the negative sign because it never states to find the magnitude of the potential difference

    http://img198.imageshack.us/img198/6914/msbn.jpg [Broken]




    I don't see how ΔV is directionless because in problem 3 it takes on the negative sign


    http://img90.imageshack.us/img90/6047/pa9.gif [Broken]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 14, 2013 #2

    rude man

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    You copied one of the equations wrong: you wrote " E = - ΔV Δs. "

    Here's the story: (vectors in bold)

    the correct formula is V2 - V1 = ΔV = - Es where Δs is defined positive pointing from point 1 to point 2
    and E is the force on a 1 Coulomb charge anywhere between points 1 and 2.

    BUT: the 1 coulomb charge is assumed to not affect the given E field. So, more correctly,
    E = limit as q → 0 of (force on a positive charge q)/q.

    Example: capacitor plates with positive voltage on plate 2 and negative voltage on plate 1.
    s points from plate 1 to plate 2.
    ΔV = V2 - V1 = -Es > 0.
    where Δs is positive since we're going from 1 to 2.
    So E = - ΔV/Δs which would be negative. The force on a small positive test charge q is always qE so the force is negative and points in the direction opposite to Δs.
     
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